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Harlamova29_29 [7]
3 years ago
7

if stall speed in ktas for an aircraft us 100 ktas at sea level, what is the stall speed in ktas of the aircraft at 5000 ft dens

ity altitude
Engineering
1 answer:
inysia [295]3 years ago
6 0

Answer:

my friend here justin

Explanation:

hes already taken

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A 35-ft³ rigid tank has propane at 25 psia, 540 R and is connected by a valve to another tank of 20 ft³ with propane at 40 psia,
gulaghasi [49]

Answer:

final pressure = 200KPa or 29.138psia

Explanation:

The detailed step by step calculations with appropriate conversion factors applied are as shown in the attachment.

8 0
3 years ago
Best use for a suspension bridge
babunello [35]

Answer:

over a rive or fast moving water or canyon

Explanation: you would use a suspension bridge in an area where you can't put supports down.

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3 years ago
Why cant you put a metals in microwaves, and toaster, etc <br><br> i have been told but never taught
12345 [234]
Jude be done. Die he ehe ehehevehe Ed
3 0
3 years ago
A system that is not influence anyway by the surroundings is called a)- control mass system b)- Isothermal system c)-- isolated
NARA [144]

Answer:

Isolated system

Explanation:

By definition of a closed system it means that a system that does not interact with it's surroundings in any manner

The other options are explained as under:

Isothermal system : It is a system that does not allow it's temperature to change

Control Mass system : It is a system whose mass remains conserved which means the mass entering the system equals the mass leaving the system

Open system: It is a system that allows transfer of mass and energy across it's boundary without any opposition i.e freely.

5 0
3 years ago
Refrigerant 134a enters an air conditioner compressor at 4 bar, 208C, and is compressed at steady state to 12 bar, 808C. The vol
SCORPION-xisa [38]

Answer:

heat transfer rate is -15.71 kW

Explanation:

given data

Initial pressure  = 4 bar

Final pressure  = 12 bar

volumetric flow rate = 4 m³ / min

work input to the compressor = 60 kJ per kg

solution

we use here super hated table for 4 bar and 20 degree temperature and 12 bar and 80 degree is

h1 = 262.96 kJ/kg

v1 = 0.05397 m³/kg

h2 = 310.24 kJ/kg

and here mass balance equation will be

m1  = m2

and mass flow equation is express as

m1 = \frac{A1\times V1}{v1}       .......................1

m1 = \frac{4\times \frac{1}{60}}{0.05397}  

m1 = 1.2353 kg/s

and here energy balance equation is express as

0 = Qcv - Wcv + m × [ ( h1-h2) + \frac{v1^2-v2^2}{2} + g (z1-z2) ]      ....................2

so here Qcv will be

Qcv =  m × [  \frac{Wcv}{m} + (h2-h1)  ]    ......................3

put here value and we get

Qcv =  1.2353 × [ {-60}+ (310.24-262.96) ]

Qcv =  -15.7130 kW

so here heat transfer rate is -15.71 kW

6 0
3 years ago
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